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statuscvo [17]
3 years ago
15

Please help on this one?

Physics
1 answer:
bezimeni [28]3 years ago
3 0

Using the given equation you get:

E = 1.99x10^-25 / 9.0x10^-6

Divide 1.99 by 9.0: 1.99/9.0 = 0.22

For the scientific notation, when dividing subtract the two exponents:

25 -6 = 19

So you now have 0.22 x 10^-19

Now you need to change the 0.22 to be in scientific notation form:

2.2 x 10^-20

The answer is B.

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Compare blue and red light from the visible spectrum. which has: the longer wavelength? the greater frequency? the greater energ
Hitman42 [59]
1) By looking at the table of the visible spectrum, we see that blue light has a wavelength in the range [450-490 nm], while red light has wavelength in the range [620-750 nm]. Therefore, red light has longer wavelength than blue light.

2) The frequency f of an electromagnetic wave is related to its wavelength \lambda by the formula
f= \frac{c}{\lambda}
where c is the speed of light. We see that the frequency is inversely proportional to the wavelength, so the shorter the wavelength, the greater the frequency. In this case, blue light has shorter wavelength than red light, so blue light has greater frequency than red light.

3) The energy of the photons of an electromagnetic wave is given by
E=hf
where h is the Planck constant and f is the frequency. We see that the energy is directly proportional to the frequency, so the greater the frequency, the greater the energy. In this problem, blue light has greater frequency than red light, so blue light has also greater energy than red light.
6 0
3 years ago
A wire carries a current of 0.66 A. This wire makes an angle of 58° with respect to a magnetic field of magnitude 5.50 10-5 T. T
Oliga [24]

Answer:

The length of the wire is "1.93 m".

Explanation:

Given:

Current,

I = 0.66 A

Angle,

\Theta = 58°

Magnetic field of magnitude,

F = 5.5\times 10^{-5} N

The length of the wire will be:

⇒ F = B\times I\times L\times Sin \Theta

or,

⇒ L = \frac{F}{B\times I\times Sin \theta}

By putting the values, we get

       =\frac{5.5\times 10^{-5}}{0.66\times 5.1\times 10^{-5}\times Sin \ 58^{\circ}}

       =\frac{5.5}{2.828}

       =1.93 \ m

6 0
3 years ago
A street musician sounds the A string of his violin, producing a tone of 440 Hz, What frequency does a bicyclist hear as he a) a
DanielleElmas [232]

Answer:

a) f_o=454.11Hz

b)f_o=425.89Hz

Explanation:

Let´s use Doppler effect, in order to calculate the observed frequency by the byciclist. The Doppler effect equation for a general case is given by:

f_o=\frac{v\pm v_o}{v\pm v_s} *f_s

where:

f_o=Observed\hspace{3}frequency

f_s=Actual\hspace{3}frequency

v=Speed\hspace{3}of\hspace{3}the\hspace{3}sound\hspace{3}waves

v_s=Velocity\hspace{3}of\hspace{3}the\hspace{3}source

v_o=Velocity\hspace{3}of\hspace{3}the\hspace{3}observer

Now let's consider the next cases:

+v_o\hspace{3}is\hspace{3}used\hspace{3}when\hspace{3}the\hspace{3}observer\hspace{3}moves\hspace{3}towards\hspace{3}the\hspace{3}source

-v_o\hspace{3}is\hspace{3}used\hspace{3}when\hspace{3}the\hspace{3}observer\hspace{3}moves\hspace{3}away\hspace{3}from\hspace{3}the\hspace{3}source

-v_s\hspace{3}is\hspace{3}used\hspace{3}when\hspace{3}the\hspace{3}source\hspace{3}moves\hspace{3}towards\hspace{3}the\hspace{3}observer

+v_s\hspace{3}is\hspace{3}used\hspace{3}when\hspace{3}the\hspace{3}source\hspace{3}moves\hspace{3}away\hspace{3}from\hspace{3}the\hspace{3}observer

The data provided by the problem is:

f_s=440Hz\\v_o=11m/s

The problem don't give us aditional information about the medium, so let's assume the medium is the air, so the speed of sound in air is:

v=343m/s

Now, in the first case the observer alone is in motion towards to the source, hence:

f_o=\frac{v+v_o}{v}*f_s=\frac{343+11}{343} *440=454.1107872Hz

Finally, in the second case the observer alone is in motion away from the source, so:

f_o=\frac{v-v_o}{v}*f_s=\frac{343-11}{343} *425.8892128Hz

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